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A318489
Number of steps to reach a lower number than starting value in 7x+-1 problem, or 0 if never reached.
1
0, 1, 12, 1, 8, 1, 4, 1, 4, 1, 42, 1, 8, 1, 4, 1, 4, 1, 23, 1, 20, 1, 4, 1, 4, 1, 12, 1, 16, 1, 4, 1, 4, 1, 282, 1, 12, 1, 4, 1, 4, 1, 229, 1, 50, 1, 4, 1, 4, 1, 8, 1, 35, 1, 4, 1, 4, 1, 8, 1, 50, 1, 4, 1, 4, 1, 46, 1, 8, 1, 4, 1, 4, 1, 225, 1, 8, 1, 4, 1, 4, 1, 35, 1, 16, 1, 4, 1, 4, 1, 46, 1, 27, 1, 4, 1, 4, 1, 16
OFFSET
1,3
COMMENTS
The least positive k for which the iterate A317640^k(n) < n.
Also called the dropping time, glide, or stopping time.
a(2n) = 1.
LINKS
D. Barina, 7x+-1: Close Relative of Collatz Problem, arXiv:1807.00908 [math.NT], 2018.
EXAMPLE
a(5) = 8 because the trajectory is (5, 36, 18, 9, 64, 32, 16, 8, 4, 2, 1, ...) and the first lower number is 4. Thus 8 steps to reach the value 4 starting from the value 5.
PROG
(C)
int a(int n0) {
if( n0 == 1 ) return 0;
int s = 0;
for(int n = n0; n >= n0; s++) {
switch(n%4) {
case 1: n = 7*n+1; break;
case 3: n = 7*n-1; break;
default: n = n/2;
}
}
return s;
}
(PARI) a7(n) = {my(m=(n+2)%4-2); if(m%2, 7*n + m, n/2)};
a(n) = if (n==1, 0, my(nb=1, m=n, nm); while((nm=a7(m)) >= n, m = nm; nb++); nb); \\ Michel Marcus, Aug 28 2018
CROSSREFS
Cf. A317640 (7x+-1 function), A102419 (3x+1 equivalent).
Sequence in context: A140378 A085094 A010214 * A121985 A245839 A068329
KEYWORD
nonn,easy
AUTHOR
David Barina, Aug 27 2018
STATUS
approved